IODS course project
Marc Denojean-Mairet
Introduction to Open Data Science - Course Project
# This is a so-called "R chunk" where you can write R code.
date()## [1] "Sun Nov 28 23:21:21 2021"I am interested in this course because of my research and I am expecting that this course will be a strong introduction to data science. The course should give me enough knowledge to be able to use data science process in my research.
I learned about the course via an email sent to me by my department.
My GitHub repository can be found here.
Here is my course diary web page.
For this analysis, a data frame named learningAnalysis2014 is created. The CSV file named “learning2014.csv” is read. The data frame consist of 7 variables (gender ,Age ,attitude, deep, stra, surf, Points) and 166 observations. The data are from a survey of statistics students. The data include the global attitude of the students toward statistics and their exam points. deep”, “stra” and “surf” are combined variables by taking the mean. “attitude” was scaled based on Likert scale (1-5) by dividing the “Attitude” column by 10.
More information about the data can be found here (https://www.mv.helsinki.fi/home/kvehkala/JYTmooc/JYTOPKYS3-meta.txt)
learningAnalysis2014 <- read.csv(file = 'data/learning2014.csv')
dim(learningAnalysis2014)## [1] 166 7str(learningAnalysis2014)## 'data.frame': 166 obs. of 7 variables:
## $ gender : chr "F" "M" "F" "M" ...
## $ Age : int 53 55 49 53 49 38 50 37 37 42 ...
## $ attitude: num 3.7 3.1 2.5 3.5 3.7 3.8 3.5 2.9 3.8 2.1 ...
## $ deep : num 3.58 2.92 3.5 3.5 3.67 ...
## $ stra : num 3.38 2.75 3.62 3.12 3.62 ...
## $ surf : num 2.58 3.17 2.25 2.25 2.83 ...
## $ Points : int 25 12 24 10 22 21 21 31 24 26 ...These are plots of all the relationship among the variables. From the visualization, we see some positive and negative correlations. An interesting correlation is attitude and points. As expected, we see a negative correlation with surf and deep. The most negative correlation is with deep and points. We also see that there are more female than male students. However, from the plots, there is no good fit between gender and points and there is no strong correlations.
The summary table gives us more information about the means of the variables.
pairs(learningAnalysis2014[-1], col = "red")
library(ggplot2)
library(GGally)## Registered S3 method overwritten by 'GGally':
## method from
## +.gg ggplot2ggpairs(learningAnalysis2014, mapping = aes(col = gender, alpha = 0.3), lower = list(combo = wrap("facethist", bins = 20)))
summary(learningAnalysis2014)## gender Age attitude deep
## Length:166 Min. :17.00 Min. :1.400 Min. :1.583
## Class :character 1st Qu.:21.00 1st Qu.:2.600 1st Qu.:3.333
## Mode :character Median :22.00 Median :3.200 Median :3.667
## Mean :25.51 Mean :3.143 Mean :3.680
## 3rd Qu.:27.00 3rd Qu.:3.700 3rd Qu.:4.083
## Max. :55.00 Max. :5.000 Max. :4.917
## stra surf Points
## Min. :1.250 Min. :1.583 Min. : 7.00
## 1st Qu.:2.625 1st Qu.:2.417 1st Qu.:19.00
## Median :3.188 Median :2.833 Median :23.00
## Mean :3.121 Mean :2.787 Mean :22.72
## 3rd Qu.:3.625 3rd Qu.:3.167 3rd Qu.:27.75
## Max. :5.000 Max. :4.333 Max. :33.00I am using the following 3 variables to explain Points: ‘attitude’, ‘stra’, and ‘deep’.
ggpairs(learningAnalysis2014, lower = list(combo = wrap("facethist", bins = 20)))
my_model <- lm(Points ~ attitude + stra + deep, data = learningAnalysis2014)
summary(my_model)##
## Call:
## lm(formula = Points ~ attitude + stra + deep, data = learningAnalysis2014)
##
## Residuals:
## Min 1Q Median 3Q Max
## -17.5239 -3.4276 0.5474 3.8220 11.5112
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 11.3915 3.4077 3.343 0.00103 **
## attitude 3.5254 0.5683 6.203 4.44e-09 ***
## stra 0.9621 0.5367 1.793 0.07489 .
## deep -0.7492 0.7507 -0.998 0.31974
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.289 on 162 degrees of freedom
## Multiple R-squared: 0.2097, Adjusted R-squared: 0.195
## F-statistic: 14.33 on 3 and 162 DF, p-value: 2.521e-08From the above summary table, we see that median for the residual is 0.5474. This would suggest that it will be difficult to predict points based on attitude, stra, and deep. However, it is very difficult to predict human behaviors and 0.5 residual could be acceptable in this case.
From the coefficients table, we see that the p-value for deep is high which would mean that deep does not affect much points. stra and attitude are better to predict points.
Below is a new regression where I have removed deep.
my_model2 <- lm(Points ~ attitude + stra, data = learningAnalysis2014)
summary(my_model2)##
## Call:
## lm(formula = Points ~ attitude + stra, data = learningAnalysis2014)
##
## Residuals:
## Min 1Q Median 3Q Max
## -17.6436 -3.3113 0.5575 3.7928 10.9295
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 8.9729 2.3959 3.745 0.00025 ***
## attitude 3.4658 0.5652 6.132 6.31e-09 ***
## stra 0.9137 0.5345 1.709 0.08927 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.289 on 163 degrees of freedom
## Multiple R-squared: 0.2048, Adjusted R-squared: 0.1951
## F-statistic: 20.99 on 2 and 163 DF, p-value: 7.734e-09Based on the above summary table, we see that by removing deep, the fit was not improved and actually got worse.
Below are diagnostic plots function: Residuals vs Fitted values, Normal QQ-plot and Residuals vs Leverage. The QQ-plot shows a reasonable fit which shows good ‘normality’. The Residuals vs Fitted values plot shows that it is reasonable since it shows randomness. The Residuals vs Leverage plot shows regular error which would imply regular leverage.
my_model2 <- lm(Points ~ attitude + stra, data = learningAnalysis2014)
par(mfrow = c(2,2))
plot(my_model2, which = c(1,2,5))
Logistic regression
Read new data frame named alc.csv to studentAlc and print variable names.
#studentAlc <- read.table("~/IODS-project/IODS-project/data/pormath.csv", sep = ",", header = TRUE)
studentAlc <- read.table("~/IODS-project/IODS-project/data/alc.csv", sep = ",", header = TRUE)
dim(studentAlc)## [1] 370 35colnames(studentAlc)## [1] "school" "sex" "age" "address" "famsize"
## [6] "Pstatus" "Medu" "Fedu" "Mjob" "Fjob"
## [11] "reason" "guardian" "traveltime" "studytime" "schoolsup"
## [16] "famsup" "activities" "nursery" "higher" "internet"
## [21] "romantic" "famrel" "freetime" "goout" "Dalc"
## [26] "Walc" "health" "failures" "paid" "absences"
## [31] "G1" "G2" "G3" "alc_use" "high_use"Data Set Information:
“This data approach student achievement in secondary education of two Portuguese schools. The data attributes include student grades, demographic, social and school related features) and it was collected by using school reports and questionnaires. Two datasets are provided regarding the performance in two distinct subjects: Mathematics (mat) and Portuguese language (por). In [Cortez and Silva, 2008], the two datasets were modeled under binary/five-level classification and regression tasks. Important note: the target attribute G3 has a strong correlation with attributes G2 and G1. This occurs because G3 is the final year grade (issued at the 3rd period), while G1 and G2 correspond to the 1st and 2nd period grades. It is more difficult to predict G3 without G2 and G1, but such prediction is much more useful (see paper source for more details).”
The above information is from UCI Machine Learning Repository.
More information about the data sets can be found here:(https://archive.ics.uci.edu/ml/datasets/Student+Performance)
I chose the following 4 variables to predict high use of alcohol: G3, absences, Pstatus, and health. I have chosen these variables thinking that they would be easily available to a school without the need to survey students.
G3:
My assumption here is that low grade is an indication of high alcohol use since high use of alcohol can affect cognitive functions such as memory.
absences:
High number of absences could be an indication of high alcohol use as it would affect one’s schedule. Absenses could be due to being sick after consuming a lot of alcohol.
goout
Going out with friend a lot might raise alcohol consumption since there will be more opportunities to consume alcohol.
health:
Poor health could be an indication of high alcohol consumption. Alcohol can negatively affect physical and mental health.
These are plots of all variables
# access the tidyverse libraries tidyr, dplyr, ggplot2
library(tidyr); library(dplyr); library(ggplot2)##
## Attaching package: 'dplyr'## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, union# draw a bar plot of each variable
gather(studentAlc) %>% ggplot(aes(value)) + facet_wrap("key", scales = "free") + geom_bar()
Box plots of my chosen variables vs high-use of alcohol
# initialize a plot of high_use and G3
g1 <- ggplot(studentAlc, aes(x = high_use, y = G3, col = sex))
# define the plot as a boxplot and draw it
g1 + geom_boxplot() + ylab("grade") + ggtitle("Student final grade by alcohol consumption and sex")
# initialise a plot of high_use and absences
g2 <- ggplot(studentAlc, aes(x = high_use, y = absences, col = sex))
# define the plot as a boxplot and draw it
g2 + geom_boxplot() + ggtitle("Student absences by alcohol consumption and sex")
# initialize a plot of high_use and
g3 <- ggplot(studentAlc, aes(x = high_use, y = goout, col = sex))
# define the plot as a boxplot and draw it
g3 + geom_boxplot() + ylab("going out") + ggtitle("Student going out with friends by alcohol consumption and sex")
g4 <- ggplot(studentAlc, aes(x = high_use, y = health, col = sex))
# define the plot as a boxplot and draw it
g4 + geom_boxplot() + ylab("health") + ggtitle("Student health by alcohol consumption and sex")
Numerical exploration of my chosen variables
# produce summary statistics by grade
studentAlc %>% group_by(sex, high_use) %>% summarise(count = n(), mean_grade = mean(G3))## `summarise()` has grouped output by 'sex'. You can override using the `.groups` argument.## # A tibble: 4 x 4
## # Groups: sex [2]
## sex high_use count mean_grade
## <chr> <lgl> <int> <dbl>
## 1 F FALSE 154 11.4
## 2 F TRUE 41 11.8
## 3 M FALSE 105 12.3
## 4 M TRUE 70 10.3# produce summary statistics by absences
studentAlc %>% group_by(sex, high_use) %>% summarise(count = n(), mean_absences = mean(absences))## `summarise()` has grouped output by 'sex'. You can override using the `.groups` argument.## # A tibble: 4 x 4
## # Groups: sex [2]
## sex high_use count mean_absences
## <chr> <lgl> <int> <dbl>
## 1 F FALSE 154 4.25
## 2 F TRUE 41 6.85
## 3 M FALSE 105 2.91
## 4 M TRUE 70 6.1# produce summary statistics by going out
studentAlc %>% group_by(sex, high_use) %>% summarise(count = n(), mean_going_out = mean(goout))## `summarise()` has grouped output by 'sex'. You can override using the `.groups` argument.## # A tibble: 4 x 4
## # Groups: sex [2]
## sex high_use count mean_going_out
## <chr> <lgl> <int> <dbl>
## 1 F FALSE 154 2.95
## 2 F TRUE 41 3.39
## 3 M FALSE 105 2.70
## 4 M TRUE 70 3.93# produce summary statistics by health
studentAlc %>% group_by(sex, high_use) %>% summarise(count = n(), mean_health = mean(health))## `summarise()` has grouped output by 'sex'. You can override using the `.groups` argument.## # A tibble: 4 x 4
## # Groups: sex [2]
## sex high_use count mean_health
## <chr> <lgl> <int> <dbl>
## 1 F FALSE 154 3.37
## 2 F TRUE 41 3.39
## 3 M FALSE 105 3.67
## 4 M TRUE 70 3.93Interpretation of the above plots and tables
We see that grades are lower for male when high_use of alcohol is true. Female seems to be less negatively affected. Final grade is not as a strong predictor as I thought. Absences and high_use of alcohol seems to have a good correlation. Female seems to have a little more absences than male when high_use of alcohol is true. going out shows some correlation for male and female for high_use of alcohol. The more a student go out the more he or she consume alcohol. Health does not seem to have a strong correlation with high_use of alcohol. I would have thought that there would have been a stronger correlation. It is possible that the participants did not truthfully answer this question or that participants are not aware of their overall health (mental and physical). This seems to be especially true for male. Female have a broader range of answers with 25% of them describing their health below 2 (Q1 high_use = true).
Logistic regression to statistically explore the relationship between my chosen variables
For this model, high_use is the target variable and final grade, absences, going out, and health are the predictors. I did not include sex because if a school needs to predict high_use of alcohol, sex might add an unnecessary bias. For instance, male students might be watched more carefully than female students because male seems to have higher high_use of alcohol.
# find the model with glm()
m <- glm(high_use ~ G3 + absences + goout + health, data = studentAlc, family = "binomial")
# print out a summary of the model
summary(m)##
## Call:
## glm(formula = high_use ~ G3 + absences + goout + health, family = "binomial",
## data = studentAlc)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.8342 -0.7505 -0.5508 0.9357 2.3172
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -3.70363 0.79076 -4.684 2.82e-06 ***
## G3 -0.03852 0.03935 -0.979 0.327587
## absences 0.07436 0.02212 3.362 0.000773 ***
## goout 0.72459 0.11941 6.068 1.29e-09 ***
## health 0.15179 0.09209 1.648 0.099275 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 452.04 on 369 degrees of freedom
## Residual deviance: 386.07 on 365 degrees of freedom
## AIC: 396.07
##
## Number of Fisher Scoring iterations: 4# print out the coefficients of the model
coef(m)## (Intercept) G3 absences goout health
## -3.70363174 -0.03852050 0.07435996 0.72459398 0.15179099# compute odds ratios (OR)
OR <- coef(m) %>% exp
# compute confidence intervals (CI)
CI <- confint(m) %>% exp## Waiting for profiling to be done...# print out the odds ratios with their confidence intervals
cbind(OR, CI)## OR 2.5 % 97.5 %
## (Intercept) 0.0246339 0.004960425 0.1109072
## G3 0.9622120 0.890726272 1.0397748
## absences 1.0771945 1.033123265 1.1280521
## goout 2.0638929 1.642693029 2.6260422
## health 1.1639169 0.973629481 1.3981834Interpretation of the above model
From the model P-values, we see that final absences and going out (goout) are likely relevant variables to explain high_use of alcohol. However, final grade and health show that the data are providing little evidence that these variables are needed to explain high_use.
From the Odd Ratio table, we see that absences, going out, and health have an OR greater than 1. This would imply that these variables are positively associated with high_use of alcohol. Going out has a Odd Ratio of 2 showing a very strong positive association with high_use of alcohol. Final grade is close to 1 as well. This would imply that the positive association is not as strong as the other variables.
It seems that most of my chosen variables have some positive association with high_use. Therefore, they could be used to correctly predict high_use.
Predictive power of my model
# predict() the probability of high_use
probabilities <- predict(m, type = "response")
# add the predicted probabilities to 'studentAlc'
studentAlc <- mutate(studentAlc, probability = probabilities)
# use the probabilities to make a prediction of high_use
studentAlc <- mutate(studentAlc, prediction = probability > 0.5)
# see the last ten original classes, predicted probabilities, and class predictions
select(studentAlc, G3, absences, goout, health, high_use, probability, prediction) %>% tail(10)## G3 absences goout health high_use probability prediction
## 361 2 7 3 3 TRUE 0.34728425 FALSE
## 362 11 3 3 3 TRUE 0.21838164 FALSE
## 363 10 2 1 5 TRUE 0.07895958 FALSE
## 364 16 4 4 2 TRUE 0.30564438 FALSE
## 365 12 3 2 3 FALSE 0.11524638 FALSE
## 366 8 4 3 3 TRUE 0.25252304 FALSE
## 367 14 0 2 5 FALSE 0.11559977 FALSE
## 368 9 4 4 5 TRUE 0.47613178 FALSE
## 369 10 8 4 2 TRUE 0.42751451 FALSE
## 370 0 0 2 5 FALSE 0.18309931 FALSE# tabulate the target variable versus the predictions
table(high_use = studentAlc$high_use, prediction = studentAlc$prediction)## prediction
## high_use FALSE TRUE
## FALSE 233 26
## TRUE 65 46# initialize a plot of 'high_use' versus 'probability' in 'studentAlc'
g <- ggplot(studentAlc, aes(x = probability, y = high_use, col = prediction))
# define the geom as points and draw the plot
g + geom_point()
# tabulate the target variable versus the predictions
table(high_use = studentAlc$high_use, prediction = studentAlc$prediction) %>% prop.table %>% addmargins## prediction
## high_use FALSE TRUE Sum
## FALSE 0.62972973 0.07027027 0.70000000
## TRUE 0.17567568 0.12432432 0.30000000
## Sum 0.80540541 0.19459459 1.00000000# define a loss function (average prediction error)
loss_func <- function(class, prob) {
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}
# call loss_func to compute the average number of wrong predictions in the (training) data
loss_func(class = studentAlc$high_use, prob = studentAlc$probability)## [1] 0.2459459The goal of a loss function is to get a small number as possible. The loss function here is around 0.25. Therefore, the model does not predict correctly 25% of the time. The model perform better than the simple guessing strategy.
10-fold cross-validation
Bonus
# K-fold cross-validation
library(boot)
cv <- cv.glm(data = studentAlc, cost = loss_func, glmfit = m, K = 10)
# average number of wrong predictions in the cross validation
cv$delta[1]## [1] 0.2486486The result of the cross-validation (K=10) is similar to the previous result of the loss function. My model seems to produce a similar result as the model found in DataCamp.
Super Bonus
Many variables
# New model with many variables
m <- glm(high_use ~ G3 + absences + goout + health + studytime + failures + freetime + famrel, data = studentAlc, family = "binomial")
# compute odds ratios (OR)
OR <- coef(m) %>% exp
# compute confidence intervals (CI)
CI <- confint(m) %>% exp## Waiting for profiling to be done...# print out the odds ratios with their confidence intervals
cbind(OR, CI)## OR 2.5 % 97.5 %
## (Intercept) 0.1112482 0.01462431 0.7969864
## G3 1.0004152 0.91884143 1.0909553
## absences 1.0707813 1.02620258 1.1197948
## goout 2.0410506 1.59736372 2.6461037
## health 1.1744770 0.97382116 1.4240680
## studytime 0.6138411 0.43227534 0.8559677
## failures 1.3512750 0.85082981 2.1704783
## freetime 1.1869927 0.89658690 1.5765989
## famrel 0.6655703 0.49995249 0.8817370# predict() the probability of high_use
probabilities <- predict(m, type = "response")
# add the predicted probabilities to 'studentAlc'
studentAlc <- mutate(studentAlc, probability = probabilities)
# use the probabilities to make a prediction of high_use
studentAlc <- mutate(studentAlc, prediction = probability > 0.5)
# define a loss function (average prediction error)
loss_func <- function(class, prob) {
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}
# call loss_func to compute the average number of wrong predictions in the (training) data
loss_func(class = studentAlc$high_use, prob = studentAlc$probability)## [1] 0.2324324# K-fold cross-validation
library(boot)
cv <- cv.glm(data = studentAlc, cost = loss_func, glmfit = m, K = 10)
# average number of wrong predictions in the cross validation
cv$delta[1]## [1] 0.2405405New model with 2 variable only
# New model with 2 variables
m <- glm(high_use ~ absences + goout, data = studentAlc, family = "binomial")
# compute odds ratios (OR)
OR <- coef(m) %>% exp
# compute confidence intervals (CI)
CI <- confint(m) %>% exp## Waiting for profiling to be done...# print out the odds ratios with their confidence intervals
cbind(OR, CI)## OR 2.5 % 97.5 %
## (Intercept) 0.02644199 0.01082308 0.06045431
## absences 1.07930582 1.03477610 1.13034717
## goout 2.08298255 1.66258410 2.64170298# predict() the probability of high_use
probabilities <- predict(m, type = "response")
# add the predicted probabilities to 'studentAlc'
studentAlc <- mutate(studentAlc, probability = probabilities)
# use the probabilities to make a prediction of high_use
studentAlc <- mutate(studentAlc, prediction = probability > 0.5)
# define a loss function (average prediction error)
loss_func <- function(class, prob) {
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}
# call loss_func to compute the average number of wrong predictions in the (training) data
loss_func(class = studentAlc$high_use, prob = studentAlc$probability)## [1] 0.2378378# K-fold cross-validation
library(boot)
cv <- cv.glm(data = studentAlc, cost = loss_func, glmfit = m, K = 10)
# average number of wrong predictions in the cross validation
cv$delta[1]## [1] 0.2432432It seems that the result is not greatly improved when going from many variables to only 2 variables.
Chapter 4: Clustering and Classification
Part 2: explore the structure and the dimensions of the data
Loading the Boston data.
# set plots size
knitr::opts_chunk$set(fig.width=16, fig.height=10)
#code from DataCamp.
#access the MASS package
library (dplyr)
library(MASS)
library(corrplot)
library(tidyr)
#load the data
data("Boston")
#explore the dataset
str(Boston)## 'data.frame': 506 obs. of 14 variables:
## $ crim : num 0.00632 0.02731 0.02729 0.03237 0.06905 ...
## $ zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
## $ indus : num 2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
## $ chas : int 0 0 0 0 0 0 0 0 0 0 ...
## $ nox : num 0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
## $ rm : num 6.58 6.42 7.18 7 7.15 ...
## $ age : num 65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
## $ dis : num 4.09 4.97 4.97 6.06 6.06 ...
## $ rad : int 1 2 2 3 3 3 5 5 5 5 ...
## $ tax : num 296 242 242 222 222 222 311 311 311 311 ...
## $ ptratio: num 15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
## $ black : num 397 397 393 395 397 ...
## $ lstat : num 4.98 9.14 4.03 2.94 5.33 ...
## $ medv : num 24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...dim(Boston)## [1] 506 14summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08205 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00Description:
The Boston data set represent the housing values in Suburbs of Boston.
The Boston data frame has 506 rows and 14 columns.
This data frame contains the following columns:
crim = per capita crime rate by town
zn = proportion of residential land zoned for lots over 25,000 sq.ft
indus = proportion of non-retail business acres per town
chas = Charles River dummy variable (= 1 if tract bounds river; 0 otherwise)
nox = nitrogen oxides concentration (parts per 10 million)
rm = average number of rooms per dwelling
age = proportion of owner-occupied units built prior to 1940
dis = weighted mean of distances to five Boston employment centres
rad = index of accessibility to radial highways
tax = full-value property-tax rate per $10,000
ptratio = pupil-teacher ratio by town
black = 1000(Bk - 0.63)^2 where Bk is the proportion of blacks by town
lstat = lower status of the population (percent)
medv = median value of owner-occupied homes in $1000s
Source:
Harrison, D. and Rubinfeld, D.L. (1978) Hedonic prices and the demand for clean air. J. Environ. Economics and Management 5, 81–102.
Belsley D.A., Kuh, E. and Welsch, R.E. (1980) Regression Diagnostics. Identifying Influential Data and Sources of Collinearity. New York: Wiley.
Part 3: graphical overview of the data
#plot matrix of the variables
pairs(Boston, gap=1/30)
# MASS, corrplot, tidyr and Boston dataset are available
# calculate the correlation matrix and round it
cor_matrix<-cor(Boston) %>% round(digits = 2)
# print the correlation matrix
cor_matrix## crim zn indus chas nox rm age dis rad tax ptratio
## crim 1.00 -0.20 0.41 -0.06 0.42 -0.22 0.35 -0.38 0.63 0.58 0.29
## zn -0.20 1.00 -0.53 -0.04 -0.52 0.31 -0.57 0.66 -0.31 -0.31 -0.39
## indus 0.41 -0.53 1.00 0.06 0.76 -0.39 0.64 -0.71 0.60 0.72 0.38
## chas -0.06 -0.04 0.06 1.00 0.09 0.09 0.09 -0.10 -0.01 -0.04 -0.12
## nox 0.42 -0.52 0.76 0.09 1.00 -0.30 0.73 -0.77 0.61 0.67 0.19
## rm -0.22 0.31 -0.39 0.09 -0.30 1.00 -0.24 0.21 -0.21 -0.29 -0.36
## age 0.35 -0.57 0.64 0.09 0.73 -0.24 1.00 -0.75 0.46 0.51 0.26
## dis -0.38 0.66 -0.71 -0.10 -0.77 0.21 -0.75 1.00 -0.49 -0.53 -0.23
## rad 0.63 -0.31 0.60 -0.01 0.61 -0.21 0.46 -0.49 1.00 0.91 0.46
## tax 0.58 -0.31 0.72 -0.04 0.67 -0.29 0.51 -0.53 0.91 1.00 0.46
## ptratio 0.29 -0.39 0.38 -0.12 0.19 -0.36 0.26 -0.23 0.46 0.46 1.00
## black -0.39 0.18 -0.36 0.05 -0.38 0.13 -0.27 0.29 -0.44 -0.44 -0.18
## lstat 0.46 -0.41 0.60 -0.05 0.59 -0.61 0.60 -0.50 0.49 0.54 0.37
## medv -0.39 0.36 -0.48 0.18 -0.43 0.70 -0.38 0.25 -0.38 -0.47 -0.51
## black lstat medv
## crim -0.39 0.46 -0.39
## zn 0.18 -0.41 0.36
## indus -0.36 0.60 -0.48
## chas 0.05 -0.05 0.18
## nox -0.38 0.59 -0.43
## rm 0.13 -0.61 0.70
## age -0.27 0.60 -0.38
## dis 0.29 -0.50 0.25
## rad -0.44 0.49 -0.38
## tax -0.44 0.54 -0.47
## ptratio -0.18 0.37 -0.51
## black 1.00 -0.37 0.33
## lstat -0.37 1.00 -0.74
## medv 0.33 -0.74 1.00# visualize the correlation matrix
corrplot(cor_matrix, method="circle", type="upper", cl.pos="b", tl.pos="d", tl.cex = 0.6)
corrplot(cor_matrix, method = 'number', type="upper")
Looking at the data, we see that medv has a positive correlation with rm and a negative correlation with lstat. This make sense as a house would have a higher price based on the number of rooms. As well, house located in a lower status population area would have a lower price. We also see that nox has a positive correlation with indus and age. nox has a negative correlation with dis. The higher the number of industry, the higher the emission of nitrogen oxides. Older houses would probably be mostly in old industrial areas of the city. medv has a positive correlation with crim but only at 0.33. At first glance, the value of houses is mainly due to th e number of rooms per dwelling. Taxes has a small negative correlation with medv.
Part 4: standardize the dataset
During scaling of the data, the mean is subtracted from the column and the difference is divided by the standard deviation. This is an example of the data before and after scaling.
Before:
row 1: rm = 6.575
After:
row 1: rm = 0.413262920
# center and standardize variables
boston_scaled <- scale(Boston)
# summaries of the scaled variables
summary(boston_scaled)## crim zn indus chas
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563 Min. :-0.2723
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668 1st Qu.:-0.2723
## Median :-0.390280 Median :-0.48724 Median :-0.2109 Median :-0.2723
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150 3rd Qu.:-0.2723
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202 Max. : 3.6648
## nox rm age dis
## Min. :-1.4644 Min. :-3.8764 Min. :-2.3331 Min. :-1.2658
## 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366 1st Qu.:-0.8049
## Median :-0.1441 Median :-0.1084 Median : 0.3171 Median :-0.2790
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059 3rd Qu.: 0.6617
## Max. : 2.7296 Max. : 3.5515 Max. : 1.1164 Max. : 3.9566
## rad tax ptratio black
## Min. :-0.9819 Min. :-1.3127 Min. :-2.7047 Min. :-3.9033
## 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876 1st Qu.: 0.2049
## Median :-0.5225 Median :-0.4642 Median : 0.2746 Median : 0.3808
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058 3rd Qu.: 0.4332
## Max. : 1.6596 Max. : 1.7964 Max. : 1.6372 Max. : 0.4406
## lstat medv
## Min. :-1.5296 Min. :-1.9063
## 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 3.5453 Max. : 2.9865# class of the boston_scaled object
class(boston_scaled)## [1] "matrix" "array"# change the object to data frame
boston_scaled <- as.data.frame(boston_scaled)Use the quantiles as the break points in the categorical variable and divide the dataset to train and test sets
# summary of the scaled crime rate
summary(boston_scaled$crim)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -0.419367 -0.410563 -0.390280 0.000000 0.007389 9.924110# create a quantile vector of crim and print it
bins <- quantile(boston_scaled$crim)
bins## 0% 25% 50% 75% 100%
## -0.419366929 -0.410563278 -0.390280295 0.007389247 9.924109610# create a categorical variable 'crime'
crime <- cut(boston_scaled$crim, breaks = bins, include.lowest = TRUE, labels = c("low", "med_low", "med_high", "high"))
# look at the table of the new factor crime
table(crime)## crime
## low med_low med_high high
## 127 126 126 127# remove original crim from the dataset
boston_scaled <- dplyr::select(boston_scaled, -crim)
# add the new categorical value to scaled data
boston_scaled <- data.frame(boston_scaled, crime)
# number of rows in the Boston dataset
n <- nrow(boston_scaled)
# choose randomly 80% of the rows
ind <- sample(n, size = n * 0.8)
# create train set
train <- boston_scaled[ind,]
# create test set
test <- boston_scaled[-ind,]
# save the correct classes from test data
correct_classes <- test$crime
# remove the crime variable from test data
test <- dplyr::select(test, -crime)Now, 80% of the data belongs to the train set. Saved the crime categories from the test set and removed the categorical crime variable from the test dataset.
Part 5: linear discriminant analysis
# linear discriminant analysis
lda.fit <- lda(crime ~ ., data = train)
# print the lda.fit object
lda.fit## Call:
## lda(crime ~ ., data = train)
##
## Prior probabilities of groups:
## low med_low med_high high
## 0.2549505 0.2425743 0.2574257 0.2450495
##
## Group means:
## zn indus chas nox rm age
## low 0.98203023 -0.9187280 -0.11943197 -0.8708372 0.45007397 -0.8840706
## med_low -0.07443837 -0.3464662 0.04906687 -0.5935372 -0.08053229 -0.3711827
## med_high -0.38746839 0.1745888 0.18195173 0.3850806 0.05872308 0.4487168
## high -0.48724019 1.0149946 -0.03371693 1.0545943 -0.44433878 0.7922113
## dis rad tax ptratio black lstat
## low 0.8724922 -0.6897369 -0.7348459 -0.43688160 0.3854282 -0.75834571
## med_low 0.4054215 -0.5494382 -0.5042940 -0.05063187 0.3199215 -0.17655895
## med_high -0.4030229 -0.4551224 -0.3404673 -0.28236407 0.1267776 0.01741206
## high -0.8396732 1.6596029 1.5294129 0.80577843 -0.7169951 0.91674504
## medv
## low 0.53533203
## med_low 0.01906654
## med_high 0.17960100
## high -0.72162353
##
## Coefficients of linear discriminants:
## LD1 LD2 LD3
## zn 0.1383955306 0.61566861 -1.07047335
## indus 0.0100457397 -0.21060477 0.28387636
## chas -0.0005458165 -0.02078708 0.15349173
## nox 0.2972943469 -0.77879828 -1.29751011
## rm -0.0204874744 0.02424114 -0.05085729
## age 0.1649926050 -0.39371511 -0.15174986
## dis -0.1966451918 -0.20987473 0.29576504
## rad 3.9278129789 0.95425484 -0.10353428
## tax 0.1603306278 -0.03634755 0.55987733
## ptratio 0.1836077744 0.01976100 -0.29747461
## black -0.1185806845 0.01321488 0.07933723
## lstat 0.0898866328 -0.17632974 0.34467425
## medv 0.0247430994 -0.41434648 -0.25297113
##
## Proportion of trace:
## LD1 LD2 LD3
## 0.9627 0.0286 0.0087# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "orange", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}
# target classes as numeric
classes <- as.numeric(train$crime)
# plot the lda results
plot(lda.fit, dimen = 2, col = classes, pch = classes)
lda.arrows(lda.fit, myscale = 1)
Part 6: Save the crime categories
# predict classes with test data
lda.pred <- predict(lda.fit, newdata = test)
# cross tabulate the results
table(correct = correct_classes, predicted = lda.pred$class)## predicted
## correct low med_low med_high high
## low 14 10 0 0
## med_low 3 14 11 0
## med_high 1 6 12 3
## high 0 0 1 27The best predictions are for high. Also, the model does not predict very well med_high. We can see in the LD plot that med_high and med_low are in the same side. The model predicts high correctly because the distance between high and low, med_low, and med_high is large.
Part 7: Reload the Boston dataset and standardize the dataset
# load MASS and Boston
library(MASS)
data('Boston')
# scale data
Boston = as.data.frame(scale(Boston))
# euclidean distance matrix
dist_eu <- dist(Boston)
# look at the summary of the distances
summary(dist_eu)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.1343 3.4625 4.8241 4.9111 6.1863 14.3970# manhattan distance matrix
dist_man <- dist(Boston, method = 'manhattan')
# look at the summary of the distances
summary(dist_man)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.2662 8.4832 12.6090 13.5488 17.7568 48.8618cluster with centers = 3
# k-means clustering
km <-kmeans(Boston, centers = 3)
# plot the Boston dataset with clusters
pairs(Boston, col = km$cluster)
### cluster with centers = 4
# k-means clustering
km <-kmeans(Boston, centers = 4)
# plot the Boston dataset with clusters
pairs(Boston, col = km$cluster)
### cluster with centers = 2
# k-means clustering
km <-kmeans(Boston, centers = 2)
# plot the Boston dataset with clusters
pairs(Boston, col = km$cluster)
model_predictors <- dplyr::select(train, -crime)
# check the dimensions
dim(model_predictors)## [1] 404 13dim(lda.fit$scaling)## [1] 13 3# matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)
#install.packages("plotly")
library(plotly)
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers')